# Left and Right Inverses of Mapping are Inverse Mapping/Proof 3

## Theorem

Let $f: S \to T$ be a mapping such that:

$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$

Then:

$g_1 = g_2 = f^{-1}$

where $f^{-1}$ is the inverse of $f$.

## Proof

Because Composition of Mappings is Associative, brackets do not need to be used.

 $\text {(1)}: \quad$ $\ds g_1 \circ f$ $=$ $\ds I_S$ $\ds \leadsto \ \$ $\ds g_1 \circ f \circ g_2$ $=$ $\ds I_S \circ g_2$ $\ds$ $=$ $\ds g_2$ Definition of Identity Mapping

 $\text {(2)}: \quad$ $\ds f \circ g_2$ $=$ $\ds I_T$ $\ds \leadsto \ \$ $\ds g_1 \circ f \circ g_2$ $=$ $\ds g_1 \circ I_T$ $\ds$ $=$ $\ds g_1$ Definition of Identity Mapping

Thus $g_1 = g_2$.

Now suppose there exists $g_3: T \to S: g_3 \circ f = I_S$.

By the same argument as above, $g_3 = g_2$.

This means that $g_1 (= g_3)$ is the only left inverse of $f$.

Similarly, suppose there exists $g_4: T \to S: f \circ g_4 = I_T$.

By the same argument as above, $g_4 = g_1$.

This means that $g_2 (= g_4)$ is the only right inverse of $f$.

So $g_1 = g_2 = g_3 = g_4$ are all the same.

By Composite of Bijection with Inverse is Identity Mapping, it follows that this unique mapping is the inverse $f^{-1}$.

$\blacksquare$